**Confession: I still haven’t figured out how to use twitter.** (Feel free to follow me @mrchasemath, though!) I always feel like I’m drinking from a fire hose when I get on the site–I can’t keep up with the twitter feed, so I don’t even try.

But *when I do*, I love seeing what people are posting. Here’s a great math problem from James Tanton. He always has such interesting problems!

Starting with zero I flip a coin. If H I add 1, if T I add 2. Repeat. What is probability I will see the number ten in my running sum?

— James Tanton (@jamestanton) April 27, 2014

Feel free to work it out yourself. It’s a fun problem! Here are my tweets that answer the question (can you follow my work?):

@jamestanton In general, the recurrence is p(n)=½p(n-1)+½p(n-2) with initial p(0)=1, p(1)=½. Solving gives p(n)=⅓(-½)^n+⅔ and p(10)=683/1024

— John Chase (@mrchasemath) April 27, 2014

@jamestanton @JosephLillo notice from the explicit formula that the probability of seeing the sum n converges to ⅔ as n → ∞ — John Chase (@mrchasemath) April 27, 2014

**It’s hard to do math with 140 characters! :-)**

Here’s his follow-up question which has still gone unanswered. My approach to the first problem won’t work here, and I want to avoid brute-forcing it. (Reminds me of my last post!) Any ideas?

Starting with zero, I roll a die and add on value I see, and repeat, to create a running sum. What is probability 7 appears in that sum? 13?

— James Tanton (@jamestanton) April 28, 2014

Let us know in the comments…or tweet @jamestanton!